Modifying a subset of rows in a pandas dataframe
Solution 1:
Use .loc
for label based indexing:
df.loc[df.A==0, 'B'] = np.nan
The df.A==0
expression creates a boolean series that indexes the rows, 'B'
selects the column. You can also use this to transform a subset of a column, e.g.:
df.loc[df.A==0, 'B'] = df.loc[df.A==0, 'B'] / 2
I don't know enough about pandas internals to know exactly why that works, but the basic issue is that sometimes indexing into a DataFrame returns a copy of the result, and sometimes it returns a view on the original object. According to documentation here, this behavior depends on the underlying numpy behavior. I've found that accessing everything in one operation (rather than [one][two]) is more likely to work for setting.
Solution 2:
Here is from pandas docs on advanced indexing:
The section will explain exactly what you need! Turns out df.loc
(as .ix has been deprecated -- as many have pointed out below) can be used for cool slicing/dicing of a dataframe. And. It can also be used to set things.
df.loc[selection criteria, columns I want] = value
So Bren's answer is saying 'find me all the places where df.A == 0
, select column B
and set it to np.nan
'
Solution 3:
Starting from pandas 0.20 ix is deprecated. The right way is to use df.loc
here is a working example
>>> import pandas as pd
>>> import numpy as np
>>> df = pd.DataFrame({"A":[0,1,0], "B":[2,0,5]}, columns=list('AB'))
>>> df.loc[df.A == 0, 'B'] = np.nan
>>> df
A B
0 0 NaN
1 1 0
2 0 NaN
>>>
Explanation:
As explained in the doc here, .loc
is primarily label based, but may also be used with a boolean array.
So, what we are doing above is applying df.loc[row_index, column_index]
by:
- Exploiting the fact that
loc
can take a boolean array as a mask that tells pandas which subset of rows we want to change inrow_index
- Exploiting the fact
loc
is also label based to select the column using the label'B'
in thecolumn_index
We can use logical, condition or any operation that returns a series of booleans to construct the array of booleans. In the above example, we want any rows
that contain a 0
, for that we can use df.A == 0
, as you can see in the example below, this returns a series of booleans.
>>> df = pd.DataFrame({"A":[0,1,0], "B":[2,0,5]}, columns=list('AB'))
>>> df
A B
0 0 2
1 1 0
2 0 5
>>> df.A == 0
0 True
1 False
2 True
Name: A, dtype: bool
>>>
Then, we use the above array of booleans to select and modify the necessary rows:
>>> df.loc[df.A == 0, 'B'] = np.nan
>>> df
A B
0 0 NaN
1 1 0
2 0 NaN
For more information check the advanced indexing documentation here.